A) \[\frac{22}{7}\]
B) \[\frac{6}{7}\]
C) \[-\frac{6}{7}\]
D) None of these
Correct Answer: B
Solution :
We have, \[x={{t}^{2}}+3t-8\] \[\Rightarrow \] \[\frac{dx}{dt}=2t+3\] and \[y=2{{t}^{2}}-2t-5\] \[\Rightarrow \] \[\frac{dy}{dt}=4t-2\] Now, \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t-2}{2t+3}\] Now, for\[x=2\]and\[y=-1,\]we have \[{{t}^{2}}+3t-8=2\]and \[2{{t}^{2}}-2t-5=-1\] \[\Rightarrow \] \[{{t}^{2}}+3t-10=0\] and \[{{t}^{2}}-t-2=0\] Solving these two equations, we get \[t=2\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(2,-1)}}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}\]You need to login to perform this action.
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